Difference between revisions of "1995 AIME Problems/Problem 2"

Revision as of 18:15, 11 October 2018

Problem

Find the last three digits of the product of the positive roots of $\sqrt{1995}x^{\log_{1995}x}=x^2$ .

Solution 1

Taking the $\log_{1995}$ (logarithm) of both sides and then moving to one side yields the quadratic equation $2(\log_{1995}x)^2 - 4(\log_{1995}x) + 1 = 0$ . Applying the quadratic formula yields that $\log_{1995}x = 1 \pm \frac{\sqrt{2}}{2}$ . Thus, the product of the two roots (both of which are positive) is $1995^{1+\sqrt{2}/2} \cdot 1995^{1 - \sqrt{2}/2} = 1995^2$ , making the solution $(2000-5)^2 \equiv \boxed{025} \pmod{1000}$ .

Solution 2

Instead of taking $\log_{1995}$ , we take $\log_x$ of both sides and simplify:

$\log_x(\sqrt{1995}x^{\log_{1995}x})=\log_x(x^{2})$

$\log_x\sqrt{1995}+\log_x x^{\log_{1995}x}=2$

$\dfrac{1}{2} \log_x 1995 + \log_{1995} x = 2$

We know that $\log_x 1995$ and $\log_{1995} x$ are reciprocals, so let $a=\log_{1995} x$ . Then we have $\dfrac{1}{2}\left(\dfrac{1}{a}\right) + a = 2$ . Multiplying by $2a$ and simplifying gives us $2a^2-4a+1=0$ , as shown above.

Because $a=\log_{1995} x$ , $x=1995^a$ . By the quadratic formula, the two roots of our equation are $a=\frac{2\pm\sqrt2}{2}$ . This means our two roots in terms of $x$ are $1995^\frac{2+\sqrt2}{2}$ and $1995^\frac{2-\sqrt2}{2}.$ Multiplying these gives $1995^2$

$1995^2\pmod{1000}\equiv 995^2\pmod{1000}\equiv (-5)^2\pmod{1000}\equiv 25\pmod{1000}$ , so our answer is $\boxed{025}$ .

Solution 3

Let $y=\log_{1995}x$ . Rewriting the equation in terms of $y$ , we have $\sqrt{1995}\left(1995^y\right)^y=1995^{2y}$ $1995^{y^2+\frac{1}{2}}=1995^{2y}$ $y^2+\frac{1}{2}=2y$ $2y^2-4y+1=0$

\[y=\frac{4\pm\sqrt{16-\left(4\right)\left(2\right)\left(1\right)}}{4}=\frac{4\pm\sqrt{8}}{4}=\frac{2\pm\sqrt{8}}{2}={1\pm\sqrt{2}}{\] (Error compiling LaTeX. Unknown error_msg)

Thus, the product of the positive roots is $\left(1995^{\frac{2+\sqrt{8}}{2}}\right)\left(1995^{\frac{2-\sqrt{8}}{2}}\right)=1995^2=\left(2000-5\right)^2$ , so the last three digits are $\boxed{025}$ .

@@ Line 27: / Line 27: @@
 <cmath> y^2+\frac{1}{2}=2y</cmath>
 <cmath> 2y^2-4y+1=0</cmath>
-<cmath> y=\frac{4\pm\sqrt{16-\left(4\right)\left(2\right)\left(1\right)}}{4}=\frac{4\pm\sqrt{8}}{4}=\frac{2\pm\sqrt{8}}{2}</cmath>
+<cmath> y=\frac{4\pm\sqrt{16-\left(4\right)\left(2\right)\left(1\right)}}{4}=\frac{4\pm\sqrt{8}}{4}=\frac{2\pm\sqrt{8}}{2}={1\pm\sqrt{2}}{</cmath>
 Thus, the product of the positive roots is <math>\left(1995^{\frac{2+\sqrt{8}}{2}}\right)\left(1995^{\frac{2-\sqrt{8}}{2}}\right)=1995^2=\left(2000-5\right)^2</math>, so the last three digits are <math>\boxed{025}</math>.

1995 AIME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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